NCERT Solutions for Class 11 Physics Chapter 6: System of Particles and Rotational Motion

Concept. The centre of mass of a body is the mass-weighted average position of its particles, R = (1/M)Σm_iri. For a body of uniform mass density with a centre of symmetry, every particle at r has a partner at −r (measured from the centre of symmetry), so the sum Σm_iri vanishes about that centre: the CM is at the geometric centre.

1. The geometric centre of each of these symmetrical bodies is the CM, because the distribution of mass is symmetric about it.
2. Does the CM necessarily lie inside the body? No. It lies within the body for solid bodies of regular shape, but not for bodies of special shapes: a ring or circular hoop (CM at the centre, in the hole), a hollow sphere or hollow cylinder (CM at the centre of the hollow), a horse-shoe (U-shaped) magnet, a boomerang, a bent rod or an L-shaped bracket may have the CM outside the material.
3. The CM is a mathematical point characterising the distribution of mass; it behaves as if all the mass were concentrated there (Section 6.3).

Examples of a CM outside the body. A ring; a hollow pipe; a bangle; a crescent; a banana (a curved body: the CM lies in the air on the concave side). The high-jumper’s ‘Fosbury flop’ technique makes use of this fact: by arching the body over the bar the CM can pass under the bar even though the body goes over it.

Extra depth. The CM of a symmetric body of non-uniform density, such as a cricket ball with a heavier side, is not at the geometric centre. For a body in a uniform gravitational field the CM coincides with the centre of gravity (CG): the point where the body can be balanced. A ring can be balanced on a finger only at points on its circumference: its CM/CG at the centre in space cannot be supported directly. The centre of mass of a system of two particles lies on the line joining them, nearer to the heavier one (Fig. 6.6).

Conclusion: (i) centre; (ii) mid-point of the axis; (iii) centre of the ring; (iv) intersection of the diagonals. No: e.g. a ring or hollow sphere has its CM outside the material.

Concept. Treat the molecule as two point masses at the nuclei, a hydrogen atom of mass m and a chlorine atom of mass 35.5 m, separated by 1.27 Å. The CM of two particles lies on the line joining them at a position determined by X = (m_1x1 + m_2x2)/(m₁ + m₂). Choose the origin conveniently at the chlorine nucleus.

1. Let the origin be at the chlorine nucleus: x_Cl = 0 (mass 35.5 m); x_H = 1.27 Å (mass m).
2. X = (35.5 m × 0 + m × 1.27)/(35.5 m + m) = 1.27/36.5 Å = 0.035 Å.
3. Hence the CM is located on the line joining the two nuclei at about 0.035 Å (3.5 × 10⁻¹² m) from the chlorine nucleus, i.e. 1.235 Å from the hydrogen nucleus (the CM is very close to the Cl nucleus, towards the H atom).

Rule: the CM divides the line joining the two masses in the inverse ratio of the masses: x_Cl : x_H = m_H : m_Cl = 1 : 35.5. The lighter atom is therefore much farther from the CM.

Extra depth. This explains why in the HCl molecule it is essentially the hydrogen atom that moves when the molecule vibrates or rotates: the heavy chlorine atom stays almost at the CM. This idea is important in molecular spectroscopy: the effective mass (reduced mass) of the HCl molecule is nearly equal to the mass of the hydrogen atom: μ = m_1m2/(m₁ + m₂) = 35.5m/36.5 = 0.97 m. The moment of inertia of the molecule about the CM is I = m_1x1² + m_2x2² = μr² = 0.97 × (1.67 × 10⁻²⁷ kg) × (1.27 × 10⁻¹⁰)² = 2.6 × 10⁻⁴⁷ kg m².

Conclusion: The CM of the HCl molecule is on the line joining the nuclei, about 0.035 Å from the chlorine nucleus (towards hydrogen).

Concept. The system trolley + child is acted upon by no external horizontal force (the floor is smooth: no friction) and the vertical forces (weights and normal reaction) cancel. The forces exerted by the child on the trolley (when running) and by the trolley on the child are internal forces (equal and opposite) and cannot change the motion of the CM of the system (M A = F_ext = 0).

1. External force on the system (horizontal) = 0.
2. Therefore the acceleration of the CM is zero: A = F_ext/M = 0.
3. So the velocity of the CM remains constant: it was V initially (the trolley and child moved together at V), hence it remains V however the child runs about on the trolley.

Detail. If the child runs forward (relative to the trolley) the trolley slows down (momentum of the trolley decreases because the child pushes it backward), and when the child stops running the original common velocity V is restored (total momentum conserved): (M + m)V = MV_T + mV_C, with V_T and V_C changing but the weighted average staying V.

Extra depth. This example shows the power of the CM concept: the complicated motions of the parts do not change the CM motion in the absence of an external force. The same is true for a person walking on a raft in still water (the CM of person + raft stays fixed in the horizontal direction): if he walks to the right the raft moves to the left. Fireworks exploding in the air, a rocket in free space, and a boat crew all obey the same rule: the CM follows the path determined by the external forces only.

Conclusion: The speed of the CM of the (trolley + child) system remains V.

Concept. The vector (cross) product has magnitude |a × b| = ab sinθ, where θ is the angle between a and b. If the two vectors are drawn from a common point O as two sides of a triangle, we can compare the area with ab sinθ.

1. Draw the vectors a = OP and b = OQ from O with angle θ between them; join P and Q to form triangle OPQ.
2. Take OP (length a) as the base. Drop a perpendicular from Q to the line OP: its length is the height h = OQ sinθ = b sinθ.
3. Area of triangle OPQ = ½ × base × height = ½ × a × b sinθ = ½ ab sinθ.
4. But |a × b| = ab sinθ. Therefore area of triangle = ½|a × b|. Proved.

Example. For a = 3î and b = 4ĵ (perpendicular): a × b = 12k̂, |a × b| = 12, so the area of the triangle is 6 (the right triangle with legs 3 and 4: ½ × 3 × 4 = 6 ✓).

Extra depth. The magnitude of the vector product is the area of the parallelogram whose adjacent sides are the two vectors; the triangle is half of it (the diagonal divides the parallelogram into two equal triangles). This geometric meaning is used in Kepler’s second law: the area swept per unit time by the radius vector, dA/dt = ½|r × v| = L/2m, is constant when the angular momentum is conserved. Using the determinant form, the area of a triangle with vertices at the origin, a and b in a plane is ½|a_xby − a_ybx|.

Conclusion: Area of the triangle = ½ a b sinθ = ½|a × b|.

Concept. Let the three vectors a, b and c be drawn from a common point O and form the three adjacent edges of a parallelepiped. The volume of a parallelepiped is (area of the base) × (height). Take the parallelogram formed by b and c as the base; its area is |b × c| and the vector b × c is perpendicular to the base.

1. The base is the parallelogram on b and c: area of base = |b × c| = bc sinφ (φ = angle between b and c).
2. The vector b × c = |b × c| n̂, where n̂ is the unit normal to the base. The height h of the parallelepiped is the perpendicular distance of the top face from the base, i.e. the component of a along n̂: h = a cosθ where θ is the angle between a and n̂ (the direction of b × c).
3. Volume V = (area of base) × (height) = |b × c| × a cosθ.
4. But the scalar product a · (b × c) = |a| |b × c| cosθ = a |b × c| cosθ, which equals V.
5. Hence |a · (b × c)| = volume of the parallelepiped. (The triple product is positive if a, b, c form a right-handed system and negative otherwise; the volume is its magnitude.)

Example. For a = î, b = ĵ, c = k̂ (edges of a unit cube): b × c = î, a · (b × c) = î·î = 1 = volume of the unit cube.

Extra depth. If the triple product is zero, the three vectors are coplanar (the parallelepiped is flat, volume zero). The cyclic symmetry a · (b × c) = b · (c × a) = c · (a × b) follows because the volume does not depend on which face is taken as the base. The scalar triple product is used in physics for the volume of the unit cell of a crystal (a · (b × c) of the lattice vectors) and for the power of a torque.

Conclusion: a · (b × c) = |b × c|(a cosθ) = (area of the base) × (height) = volume of the parallelepiped.

Concept. The angular momentum of a particle about the origin is l = r × p. Write r = xî + yĵ + zk̂ and p = p_xî + p_yĵ + p_zk̂ and expand the cross product using î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (and the reverse products with a minus sign; like unit vectors give zero).

1. l = r × p = | î ĵ k̂ ; x y z ; p_x p_y p_z |.
2. Expanding the determinant: l = (y p_z − z p_y)î + (z p_x − x p_z)ĵ + (x p_y − y p_x)k̂.
3. Hence the components are: l_x = y p_z − z p_y, l_y = z p_x − x p_z, l_z = x p_y − y p_x.
4. Motion in the x–y plane: z = 0 and p_z = 0 (no component of position or momentum along z). Then l_x = y(0) − 0 × p_y = 0; l_y = 0 × p_x − x(0) = 0; and only l_z = x p_y − y p_x survives.
5. So for a particle moving in the x–y plane l = (x p_y − y p_x)k̂, i.e. the angular momentum is directed along the z-axis, perpendicular to the plane of motion (as the right-hand rule requires for r × p, since both r and p lie in the x–y plane).

Example. A particle at r = (3, 4, 0) m with p = (2, −1, 0) kg m s⁻¹: l_z = 3(−1) − 4(2) = −11 kg m² s⁻¹, l_x = l_y = 0: the angular momentum is 11 units along the −z axis (clockwise motion seen from +z).

Extra depth. In polar form l_z = r p sinθ, which equals mvr for circular motion (θ = 90°). In three dimensions the three components are the same as the 3 independent torque/angular momentum components used in the study of rotation; l is perpendicular to both r and p, so l · r = 0 and l · p = 0 (you can check: x l_x + y l_y + z l_z = 0). A plane motion about a point in the plane has only one non-zero component, which is why in this chapter angular momentum for planar problems is treated as a scalar with a sign.

More to know — angular momentum of a particle in circular motion. For uniform circular motion in the x–y plane with radius r and angular speed ω: x = r cosωt, y = r sinωt, p_x = −mrω sinωt, p_y = mrω cosωt, so l_z = xp_y − yp_x = mr²ω(cos² + sin²) = mr²ω = mvr, constant in time and along +z for counter-clockwise motion. The same formula gives L = Iω for a rigid body. The angular momentum of a particle moving in a straight line with constant velocity is also constant (Example 6.6) and equals mv times the perpendicular distance of the line from the origin.

Conclusion: l_x = yp_z − zp_y, l_y = zp_x − xp_z, l_z = xp_y − yp_x; for planar motion in the xy-plane only l_z is non-zero.

Concept. The angular momentum of the system about a point O is L = r₁ × p₁ + r₂ × p₂, where r₁, r₂ are the position vectors of the two particles from O and p₁ = mv, p₂ = −mv are their momenta (equal in magnitude, opposite in direction). We show that L does not depend on the position of O.

1. Let the particles be at positions r₁ and r₂ relative to an arbitrary origin O. The velocities are v and −v, so p₁ = mv and p₂ = −mv.
2. L = r₁ × mv + r₂ × (−mv) = m(r₁ − r₂) × v.
3. The vector r₁ − r₂ = r₁₂ is the position vector of particle 1 relative to particle 2: it depends only on the positions of the two particles, not on the origin (any shift of the origin by a vector a changes r₁ and r₂ by the same a and leaves the difference unchanged).
4. Therefore L = m r₁₂ × v is independent of the choice of O. Its magnitude is m v (r₁₂ sinθ) = m v d, because the component of r₁₂ perpendicular to the (parallel) lines of motion is the separation d; its direction is perpendicular to the plane containing the two lines of motion.

Check with distances. If O lies between the two lines at perpendicular distances x_1 and x_2 from them (x_1 + x_2 = d), the two contributions have the same sense and add up: mvx_1 + mvx_2 = mvd. If O lies outside, at distances x and x + d from the lines, the contributions have opposite senses: mv(x + d) − mvx = mvd. The result is always mvd.

Extra depth. The result is a special case of a general theorem: the angular momentum of a system depends on the choice of the origin by an amount R × P where R is the shift and P the total momentum. Here the total momentum is p₁ + p₂ = 0, so the angular momentum is independent of the origin. The two particles form a ‘couple’-like arrangement (equal and opposite momenta along different lines), whose angular momentum, like the moment of a couple of forces (Example 6.7), is independent of the reference point.

Conclusion: The total angular momentum L = m (r₁ − r₂) × v = mvd (perpendicular to the plane) is the same about every point.

Concept. The bar is in equilibrium under three forces: its weight W (acting at the CG, a distance d from the left end), and the tensions T₁ (left string, 36.9° to the vertical) and T₂ (right string, 53.1° to the vertical). Equilibrium requires (i) ΣF_x = 0, (ii) ΣF_y = 0 (translational) and (iii) Στ = 0 (rotational), i.e. torques about any point sum to zero.

1. Use sin36.9° = 0.6, cos36.9° = 0.8, sin53.1° = 0.8, cos53.1° = 0.6.
2. Horizontal: T₁ sin36.9° = T₂ sin53.1° ⇒ 0.6 T₁ = 0.8 T₂ ⇒ T₁ = (4/3)T₂.
3. Vertical: T₁ cos36.9° + T₂ cos53.1° = W ⇒ 0.8 T₁ + 0.6 T₂ = W ⇒ 0.8(4/3)T₂ + 0.6T₂ = W ⇒ (16/15 + 9/15) T₂ = (5/3) T₂ = W ⇒ T₂ = 0.6 W and T₁ = 0.8 W.
4. Torques about the left end (taking anticlockwise positive): the weight acts downward at distance d (clockwise torque W d); the right string’s vertical component T₂ cos53.1° acts upward at distance 2 m (anticlockwise torque T₂ cos53.1° × 2); the left tension passes through the left end (zero torque) and its horizontal components act along the line of the bar (no torque about the end).
5. Rotational equilibrium: T₂ cos53.1° × 2 = W d ⇒ 0.6W × 0.6 × 2 = W d ⇒ d = 0.72 m.

Check (torques about the right end): W (2 − d) = T₁ cos36.9° × 2 ⇒ W(2 − d) = 0.8W × 0.8 × 2 = 1.28W ⇒ 2 − d = 1.28 ⇒ d = 0.72 m ✓.

Extra depth. The CG is closer to the left end (0.72 m from the left, 1.28 m from the right), as expected from the fact that the left string, nearer to the vertical, carries the larger tension (0.8W): the larger share of the weight is supported near the CG’s side. A uniform bar would have its CG at d = 1 m (the middle). The method — resolve forces, then take moments about a convenient point where unknown forces pass — is the standard approach to rigid-body equilibrium problems.

Conclusion: d = 0.72 m from the left end.

Concept. The car is in equilibrium on level ground: the total upward reaction of the ground equals the weight, and the torques about any point vanish. Let R_f be the total reaction on the two front wheels and R_b on the two back wheels. The weight acts at the CG, 1.05 m behind the front axle, i.e. 1.8 − 1.05 = 0.75 m in front of the back axle.

1. Weight W = 1800 kg × 9.8 m s⁻² = 17640 N (taking g = 9.8 m s⁻²; 1800 kg wt).
2. Translational equilibrium: R_f + R_b = W = 17640 N.
3. Rotational equilibrium (torques about the back axle): R_f × 1.8 = W × 0.75 ⇒ R_f = 17640 × 0.75/1.8 = 7350 N.
4. R_b = 17640 − 7350 = 10290 N.
5. The force on each wheel: front wheel = R_f/2 = 3675 N (≈ 375 kg wt); back wheel = R_b/2 = 5145 N (≈ 525 kg wt).

Check by torques about the front axle: R_b × 1.8 = W × 1.05 ⇒ R_b = 17640 × 1.05/1.8 = 10290 N ✓.

Extra depth. The back wheels carry more weight (58%) because the CG is nearer to the back axle (0.75 m) than to the front (1.05 m). The load on the wheels is inversely proportional to the distance of the CG from the respective axle: R_f : R_b = 0.75 : 1.05 = 5 : 7. This is why heavy loads are put towards the rear in a truck, and why braking transfers weight to the front wheels (dynamic load transfer). In the exercise the ‘1800 kg’ is a mass and the forces are weights in newtons (using g = 9.8 m s⁻²).

Conclusion: Front wheel: 3675 N; back wheel: 5145 N (g = 9.8 m s⁻²).

Concept. Rotational dynamics: τ = Iα, so the angular acceleration α = τ/I. Starting from rest, the angular speed after time t is ω = αt = τt/I. For the same torque and time, the body with the smaller moment of inertia attains the greater angular speed. Compare the moments of inertia from Table 6.1.

1. I₁ = MR² and I₂ = (2/5)MR² = 0.4 MR²; so I₂ < I₁.
2. α₁ = τ/I₁ and α₂ = τ/I₂ ⇒ α₂/α₁ = I₁/I₂ = 5/2 = 2.5.
3. After time t (starting from rest) ω = αt, so ω_sphere/ω_cylinder = 2.5.
4. Therefore the solid sphere acquires the greater angular speed (2.5 times that of the hollow cylinder).

Physical reason. In the hollow cylinder all the mass is at the maximum distance R from the axis, which gives the largest possible I for the given M and R (I = MR²). In the solid sphere much of the mass is near the axis, so I is small, and the same torque produces a large angular acceleration.

Extra depth. The moment of inertia plays the role of mass in rotation: the larger the I, the harder to start or stop the rotation. This is why a flywheel (a large-I wheel with mass concentrated at the rim) smooths out engine fluctuations, while a figure skater reduces I by pulling the arms close to the body in order to spin fast. The result ω ∝ 1/I also explains why a hollow pipe is harder to spin than a solid rod of the same mass and radius.

More to know — moments of inertia compared. For the same mass M and radius R: hollow cylinder (or ring about its axis) MR²; solid cylinder MR²/2; hollow sphere 2MR²/3; solid sphere 2MR²/5. A larger I means a smaller angular acceleration for the same torque and also a smaller kinetic energy for the same angular momentum (K = L²/2I). When bodies roll down an incline the one with the smallest I/MR² arrives first: solid sphere (0.4) beats solid cylinder (0.5), which beats hollow sphere (0.67), which beats hollow cylinder or ring (1) — a standard NEET question related to this exercise.

Conclusion: The solid sphere acquires the greater angular speed (2.5 times that of the hollow cylinder) since its moment of inertia is smaller.

Concept. For rotation about a fixed axis the kinetic energy is K = ½Iω² and the angular momentum is L = Iω. For a solid cylinder about its own axis the moment of inertia is I = ½MR² (Table 6.1).

1. Mass M = 20 kg, radius R = 0.25 m, ω = 100 rad s⁻¹.
2. Moment of inertia I = ½MR² = ½ × 20 × (0.25)² = 10 × 0.0625 = 0.625 kg m².
3. Kinetic energy of rotation K = ½Iω² = ½ × 0.625 × (100)² = ½ × 0.625 × 10⁴ = 3125 J (≈ 3.1 × 10³ J).
4. Angular momentum L = Iω = 0.625 × 100 = 62.5 J s (kg m² s⁻¹).

Relations: K = L²/(2I) = (62.5)²/(2 × 0.625) = 3125 J. The linear speed of a point on the rim is v = ωR = 100 × 0.25 = 25 m s⁻¹ (90 km h⁻¹), and a particle on the rim has a centripetal acceleration ω²R = 2500 m s⁻².

Extra depth. The energy stored in the spinning cylinder (3.1 kJ) is large relative to its size: this is how flywheels store energy (used in some buses and in energy-storage devices). If the cylinder had been hollow (I = MR² = 1.25 kg m²), at the same ω the energy would be 6250 J and L = 125 J s: twice as much, because the mass is concentrated at the rim.

More to know — rotational analogues. Every translational formula has a rotational partner: K = ½Mv² ↔ ½Iω²; p = Mv ↔ L = Iω; F = Ma ↔ τ = Iα; P = Fv ↔ τω. For the cylinder of the exercise, stopping it in 10 s needs a constant retarding torque τ = Iω/t = 0.625 × 100/10 = 6.25 N m, and the work done in stopping it equals its kinetic energy, 3125 J, which would appear as heat in the brake. Energy stored in a flywheel is proportional to Iω², so doubling the speed stores four times as much energy.

Conclusion: K = 3125 J (3.1 × 10³ J) and L = 62.5 kg m² s⁻¹.

Concept. With no external torque (frictionless turntable) the angular momentum of the child + turntable is conserved: I₁ω₁ = I₂ω₂. The child changes his moment of inertia by moving his arms, which changes the angular speed inversely. The kinetic energy K = ½Iω² is not conserved: it changes because of internal work done by the child.

(a) New angular speed

1. I₁ = I (initial), ω₁ = 40 rev/min; I₂ = (2/5)I.
2. Conservation of angular momentum: I₁ω₁ = I₂ω₂ ⇒ ω₂ = (I₁/I₂)ω₁ = (5/2) × 40 = 100 rev/min.
3. In rad s⁻¹: ω₁ = 2π × 40/60 = 4.19 rad s⁻¹; ω₂ = 10.5 rad s⁻¹.

(b) Rotational kinetic energy

1. K₁ = ½I₁ω₁², K₂ = ½I₂ω₂².
2. K₂/K₁ = (I₂/I₁)(ω₂/ω₁)² = (2/5)(5/2)² = (2/5)(25/4) = 5/2 = 2.5.
3. So the new kinetic energy is 2.5 times the initial one: it has increased (by 150%).
4. Source of the extra energy: the child does work against the ‘outward’ (centrifugal-like) tendency of the arms while pulling them in: the muscular (internal) force acting along the displacement of the arms does positive work, converting the child’s chemical (internal) energy into rotational kinetic energy. The total energy is conserved when the child’s chemical energy is counted.

General relation: K = L²/(2I); at constant L, K ∝ 1/I. Reducing I to 2/5 increases K by 5/2. The ‘work done by the child’ is 1.5 K₁.

Extra depth. The same physics governs a skater who spins faster when pulling the arms in, a diver in a tuck, the collapse of a star (a rotating cloud of gas contracting spins faster: neutron stars may rotate hundreds of times per second) and an ice dancer. Angular momentum is conserved because the net external torque is zero (the turntable is frictionless and the child’s weight is balanced by the normal reaction along the axis). A paradox only arises if one forgets that the internal forces (muscles) can do work even though they exert no net external torque.

Conclusion: (a) 100 rev/min; (b) K₂/K₁ = 2.5: the increase comes from the work done by the child (internal energy).

Concept. The tension (pull) of the rope applies a torque about the axis of the cylinder: τ = F R. The rotational equation of motion τ = Iα then gives the angular acceleration, with I = MR² for a hollow cylinder about its axis. If the rope does not slip on the cylinder, the linear acceleration of the rope equals the tangential acceleration of the rim: a = αR.

1. M = 3 kg, R = 40 cm = 0.40 m, F = 30 N.
2. Torque τ = F R = 30 × 0.40 = 12 N m.
3. Moment of inertia of the hollow cylinder (thin-walled) about its axis I = MR² = 3 × (0.40)² = 0.48 kg m².
4. Angular acceleration α = τ/I = 12/0.48 = 25 rad s⁻².
5. Linear acceleration of the rope a = αR = 25 × 0.40 = 10 m s⁻² (equal to the tangential acceleration of the rim).

Check by linear dynamics? For a hollow cylinder pulled by a rope with force F, the rope acceleration a = F R²/I = F/M = 30/3 = 10 m s⁻²: the same as for a free particle of mass M pulled by F, because for a thin-walled cylinder all the mass is at the rim, so the rotational inertia equals the translational inertia M when expressed in rim terms.

Extra depth. For a solid cylinder (I = ½MR²) the same pull would give α = 50 rad s⁻² and a = 20 m s⁻², twice as large, since its moment of inertia is half. If the cylinder were free to move as well (not fixed on an axis), the equations would change (translation + rotation). Here the cylinder is on a fixed axis, and the force and the rope are assumed ideal: the hand supplies a constant 30 N.

More to know — solid versus hollow cylinder. The acceleration of a rope wound on a cylinder fixed on an axis is a = F R²/I. With I = MR² (hollow) a = F/M = 10 m s⁻²; with I = ½MR² (solid) a = 2F/M = 20 m s⁻². If the cylinder is free to roll down while the rope end is fixed (a yo-yo), the acceleration of the centre is a = g/(1 + I/MR²): g/2 for a hollow cylinder and 2g/3 for a solid cylinder. These comparisons show how a larger moment of inertia gives a smaller angular acceleration.

Conclusion: α = 25 rad s⁻²; linear acceleration of the rope = 10 m s⁻².

Concept. The power transmitted by a torque τ to a body rotating with angular velocity ω is P = τω (the rotational analogue of P = Fv). At uniform angular speed the engine’s torque only balances the frictional torque (net torque zero, so no angular acceleration), but it continuously supplies energy at the rate τω to compensate for the heat produced by friction.

1. Torque τ = 180 N m, angular speed ω = 200 rad s⁻¹.
2. Power P = τω = 180 × 200 = 36000 W = 3.6 × 10⁴ W (36 kW).
3. Since the engine is assumed 100% efficient, this is also the power input; in horse power: 36000/746 = 48 hp.

Energy in one minute: E = Pt = 36 kW × 60 s = 2.16 MJ = 0.6 kWh; all of it appears as heat generated by friction in the bearings, since the kinetic energy of the rotor does not change at constant ω.

Extra depth. Derivation: dW = τdθ, so P = dW/dt = τdθ/dt = τω. The torque times the rotation speed is also the way vehicle power is specified: engine power (kW) = torque (N m) × 2π × (rev/min)/60000. For 1910 rev/min and 180 N m this reproduces 36 kW. At the same power a lower speed demands a higher torque, which is why gears are used.

More to know — engines and torque. Engine power P = τω, so for a fixed power the torque is inversely proportional to the speed: the same 36 kW engine delivers 180 N m at 200 rad s⁻¹ but 360 N m at 100 rad s⁻¹. Gearboxes convert high speed/low torque into low speed/high torque for climbing a hill. In an electric motor the torque is proportional to the current. The unit of torque (N m) is dimensionally the joule but torque is a vector while energy is a scalar; do not call torque ‘joule’.

Conclusion: The power required is 3.6 × 10⁴ W (36 kW).

Concept. Use the superposition (negative mass) method: the original full disc = (the remaining body) + (the hole piece). The CM of the whole disc is at its centre O; the CM of the cut-out disc is at its own centre, at a distance R/2 from O. Treat the remaining body as the whole disc minus the hole. The CG coincides with the CM for a flat body in uniform gravity.

1. Let the mass per unit area be σ. Mass of the full disc M = σπR². Mass of the removed disc (radius R/2) = σπ(R/2)² = M/4. Mass of the remaining body = M − M/4 = 3M/4.
2. Take the origin at the centre O of the original disc and the x-axis along the line from O to the centre of the hole (hole centre at x = +R/2).
3. CM of the full disc (at O) = [ (3M/4) x_rem + (M/4)(R/2) ]/M = 0 (since the full disc is symmetric about O).
4. So (3M/4) x_rem + MR/8 = 0 ⇒ x_rem = −(MR/8)/(3M/4) = −R/6.
5. Hence the CG of the remaining body is at a distance R/6 from the centre O of the original disc, on the side opposite to the hole (on the line joining O to the centre of the hole).

Check by direct moments: M(0) = (3M/4)(−R/6) + (M/4)(R/2) = −MR/8 + MR/8 = 0 ✓.

Extra depth. The method of treating a hole as a body of negative mass is a standard trick for finding the CM of bodies with cut-outs. It applies equally to moments of inertia and to gravitational and electrostatic fields. The remainder balances at a point on the side away from the hole, which makes sense: removing mass from the right side shifts the CM to the left. In the general case with a hole of radius r whose centre is at distance a from O, the shift is x = −(r² a)/(R² − r²).

Conclusion: The CG of the remaining body is at R/6 from the centre of the original disc, on the side opposite to the hole.

Concept. A uniform metre stick has its centre of gravity at the 50.0 cm mark: this is where it balances on its own. With the two coins on the 12.0 cm mark, the system balances at 45.0 cm, so the total torque of the weights about the 45.0 cm mark must vanish (principle of moments: clockwise moment = anticlockwise moment).

1. Mass of the coins m_c = 2 × 5 g = 10 g at x = 12.0 cm; let the mass of the stick be M at x = 50.0 cm (its CG).
2. Pivot (knife edge) at x = 45.0 cm. Distance of the coins from the pivot = 45.0 − 12.0 = 33.0 cm (left of the pivot); distance of the stick’s CG from the pivot = 50.0 − 45.0 = 5.0 cm (right of the pivot).
3. Moments about the pivot: 10 g × 33.0 cm = M × 5.0 cm (g cancels).
4. M = 10 × 33.0/5.0 = 66 g.

Check using the CM: CM of the system = (10 × 12 + 66 × 50)/(10 + 66) = (120 + 3300)/76 = 3420/76 = 45.0 cm ✓, which is the observed balance point.

Extra depth. This is a practical method to find the mass of an object using a balance of known masses (the principle of moments), the basis of the beam balance, the lever and the steelyard. In the experimental physics of metre bridge or metre-stick balances the knife edge is moved until the stick is balanced; the position gives the CM of the combined system. Note that the mass of the stick, 66 g, is realistic for a wooden metre stick.

More to know — the principle of moments in practice. A balance works by equating moments: m_1d1 = m_2d2. A beam balance compares masses; a steelyard uses a sliding known mass; the metre-stick method used here is an inexpensive way to weigh light objects: with the pivot at 45 cm and a 10 g load at 12 cm one finds a stick mass of 66 g. Sensitivity: moving the coins to the 6 cm mark would shift the balance point to (10 × 6 + 66 × 50)/76 = 44.2 cm; the shift of 0.8 cm is easily measurable, so a few grams can be weighed to within a gram.

Conclusion: The mass of the metre stick is 66 g.

Concept. A diatomic molecule has translational kinetic energy K_t = ½mv² and rotational kinetic energy K_r = ½Iω² about the perpendicular axis through its centre. The data give K_r = (2/3)K_t, from which ω is found.

1. m = 5.30 × 10⁻²⁶ kg, v = 500 m s⁻¹, I = 1.94 × 10⁻⁴⁶ kg m².
2. K_t = ½mv² = ½ × 5.30 × 10⁻²⁶ × (500)² = ½ × 5.30 × 10⁻²⁶ × 2.5 × 10⁵ = 6.625 × 10⁻²¹ J.
3. K_r = (2/3)K_t = (2/3) × 6.625 × 10⁻²¹ = 4.417 × 10⁻²¹ J.
4. K_r = ½Iω² ⇒ ω² = 2K_r/I = 2 × 4.417 × 10⁻²¹/1.94 × 10⁻⁴⁶ = 4.55 × 10²⁵.
5. ω = √(4.55 × 10²⁵) = 6.75 × 10¹² rad s⁻¹.

Comparison. The rotation frequency ν = ω/2π = 1.07 × 10¹² Hz (about 10¹² revolutions per second, in the far-infrared region). This is typical of molecular rotations. The angular momentum of the molecule L = Iω = 1.94 × 10⁻⁴⁶ × 6.75 × 10¹² = 1.31 × 10⁻³³ J s, which is of the order of the Planck constant h/2π (1.05 × 10⁻³⁴ J s): rotational motion at atomic level is quantised.

Extra depth. The factor 2/3 has been chosen to be consistent with the equipartition of energy in the kinetic theory of gases (Chapter 12): for a diatomic molecule at room temperature there are 3 translational and 2 rotational degrees of freedom, each with ½kT, so K_r/K_t = 2/3. This exercise therefore links mechanics to thermodynamics: the average angular velocity (6.75 × 10¹² rad s⁻¹) of an oxygen molecule at room temperature is enormous compared with everyday rotations.

Conclusion: The average angular velocity of the oxygen molecule is 6.75 × 10¹² rad s⁻¹.